Discriminant and positivity

Completed square.
$$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a}.$$

Case $\Delta<0$. Then $4ac-b^2=-\Delta>0$, and since $a>0$, the second term is positive. The first term is $\ge0$, so the sum is $>0$ for all $x$.

Case $\Delta\ge0$. Then $4ac-b^2\le0$. At $x=-\frac{b}{2a}$, the quadratic equals $\frac{4ac-b^2}{4a}\le0$, so it is not always positive.

Conclusion. $ax^2+bx+c>0$ for all $x$ iff $a>0$ and $\Delta<0$.